A uniform metal bar $70 ext{ cm}$ long and $4.00 ext{ kg}$ in mass is supported on two knife-edges

Question

A uniform metal bar $70	ext{ cm}$ long and $4.00	ext{ kg}$ in mass is supported on two knife-edges placed $10	ext{ cm}$ from each end. A $6.00	ext{ kg}$ load is suspended at $30	ext{ cm}$ from one end. What are the reaction forces at the knife-edge nearer to the load and the farther knife-edge, respectively? (Take $g = 9.8	ext{ m/s}^2$)

Professor PC Thomas Classes · Answer

Let the bar be $AB$. Center of gravity $G$ is at $35	ext{ cm}$. Weight of bar $W = 4 	imes 9.8 = 39.2	ext{ N}$ acts at $G$. Knife edges $K_1$ and $K_2$ are at $10	ext{ cm}$ and $60	ext{ cm}$ from $A$. Load $W_1 = 6 	imes 9.8 = 58.8	ext{ N}$ is at $30	ext{ cm}$. Taking moments about $G$: $R_1(25	ext{ cm}) - W_1(5	ext{ cm}) - R_2(25	ext{ cm}) = 0$. So, $R_1 - R_2 = 58.8 	imes \frac{5}{25} = 11.76	ext{ N}$. From translational equilibrium: $R_1 + R_2 = 39.2 + 58.8 = 98	ext{ N}$. Solving these gives $R_1 = 54.88	ext{ N}$ and $R_2 = 43.12	ext{ N}$.